Problem: What is the extraneous solution to these equations? $\dfrac{x^2 - 7x}{x + 3} = \dfrac{-17x - 21}{x + 3}$
Solution: Multiply both sides by $x + 3$ $ \dfrac{x^2 - 7x}{x + 3} (x + 3) = \dfrac{-17x - 21}{x + 3} (x + 3)$ $ x^2 - 7x = -17x - 21$ Subtract $-17x - 21$ from both sides: $ x^2 - 7x - (-17x - 21) = -17x - 21 - (-17x - 21)$ $ x^2 - 7x + 17x + 21 = 0$ $ x^2 + 10x + 21 = 0$ Factor the expression: $ (x + 7)(x + 3) = 0$ Therefore $x = -7$ or $x = -3$ At $x = -3$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -3$, it is an extraneous solution.